///done!~
/*
给你一个数组 items ，其中 items[i] = [typei, colori, namei] ，描述第 i 件物品的类型、颜色以及名称。

另给你一条由两个字符串 ruleKey 和 ruleValue 表示的检索规则。

如果第 i 件物品能满足下述条件之一，则认为该物品与给定的检索规则 匹配 ：

    ruleKey == "type" 且 ruleValue == typei 。
    ruleKey == "color" 且 ruleValue == colori 。
    ruleKey == "name" 且 ruleValue == namei 。

统计并返回 匹配检索规则的物品数量 。

 

示例 1：

输入：items = [["phone","blue","pixel"],["computer","silver","lenovo"],["phone","gold","iphone"]], ruleKey = "color", ruleValue = "silver"
输出：1
解释：只有一件物品匹配检索规则，这件物品是 ["computer","silver","lenovo"] 。

示例 2：

输入：items = [["phone","blue","pixel"],["computer","silver","phone"],["phone","gold","iphone"]], ruleKey = "type", ruleValue = "phone"
输出：2
解释：只有两件物品匹配检索规则，这两件物品分别是 ["phone","blue","pixel"] 和 ["phone","gold","iphone"] 。注意，["computer","silver","phone"] 未匹配检索规则。

 

提示：

    1 <= items.length <= 104
    1 <= typei.length, colori.length, namei.length, ruleValue.length <= 10
    ruleKey 等于 "type"、"color" 或 "name"
    所有字符串仅由小写字母组成

通过次数10,293
提交次数12,167
*/

/*
 * @Author:gxx
 * @QQ : 1278255275
 * @Date: 2021-04-01 14:51:35
 * @Last Modified by: gxx
 * @Last Modified time: 2021-04-01 14:51:35
 * @Description: 统计匹配检索规则的物品数量
*/
// [["phone","blue","pixel"],["computer","silver","lenovo"],["phone","gold","iphone"]]
// "color"
// "silver"

class Solution
{
public:
    int countMatches(vector<vector<string>> &items, string ruleKey, string ruleValue)
    {
        int len = items.size();
        int num = 0;
        if (ruleKey == "type") {
            for (int i = 0; i < len; i++){
                    if (items[i][0] == ruleValue)
                        num++;
            }
        }else if(ruleKey == "color") {
            for (int i = 0; i < len; i++){
                    if (items[i][1] == ruleValue)
                        num++;
            }
        }else if(ruleKey =="name"){
            for (int i = 0; i < len; i++){
                    if (items[i][2] == ruleValue)
                        num++;
            }
        }
        cout << num << endl;
        return num;
    }
};